Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = numeric_limits<int>::max();
const long long LLINF = numeric_limits<long long>::max();
const unsigned long long ULLINF = numeric_limits<unsigned long long>::max();
const double PI = acos(-1.0);
int n, H;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> H;
  double b = 1;
  vector<double> v;
  double h = H;
  for (int i = 0; i < n - 1; i++) {
    double D = 4.0 * b * b * h * h - 4.0 * h * H * b / n;
    double x1 = (2.0 * b * h + sqrt(D)) / (2.0 * b);
    double x2 = (2.0 * b * h - sqrt(D)) / (2.0 * b);
    double x;
    if (x2 < 0)
      x = x1;
    else
      x = x2;
    v.push_back(h - x);
    b = 1.0 * b * (h - x) / h;
    h -= x;
  }
  reverse((v).begin(), (v).end());
  for (auto a : v) cout << setprecision(10) << fixed << a << " ";
  return 0;
}